In the case of \(k^{(I)}=0\) , from Eq. (), we easily get \(x=0\)  and \(y=1/\left(k^{(U)}-1\right)\) , in agreement with prior work in the bi-directed setting [1]}.
For \(k^{(U)}=0\) , we have \(x=1/\left(2k^{(I)}-1\right)\)  and \(y=0\) .
In the following, we focus on the case with \(k^{(I)}\ge 1, k^{(U)}\ge 1\) .
Moving the second terms in Eq. () to the right side and multiplying the two equations, we have
\(y=\frac{-(2k^{(I)}-1)x+1}{(2k^{(I)}+k^{(U)}-1)x+k^{(U)}-1}. Substituting p_{AB}^{(I)}=p_A\left(1-q_{A|A}^{(I)}\right) and p_{AB}^{(U)}=p_A\left(1-q_{A|A}^{(U)}\right) into the first equality in Eq.~(\ref {time_scale1}), and using Eqs.~(\ref {qAA},\ref {yx}), we have{\equation ax^3+bx^2+cx+d=0, \endequation }where{\equation \begin{split}a=& 4\left(k^{(I)}\right)^2+2k^{(I)}k^{(U)}-4k^{(I)}-k^{(U)}+1, \\b=& -4\left(k^{(I)}\right)^2-8k^{(I)}k^{(U)}-2\left(k^{(U)}\right)^2+3k^{(U)}+1,\\c=& -2k^{(I)}k^{(U)}+4k^{(I)}-2\left(k^{(U)}\right)^2+5k^{(U)}-1,\\d=& k^{(U)}-1.\end{split}\endequation }According to Eq.~(\ref {time_scale1}), the solution of Eq.~(\ref {cubic}), x^*, must satisfy -1< x^*< 1 and -1< y(x^*)< 1, which gives  -\left(k^{(U)}-2\right)/\left(4k^{(I)}+k^{(U)}-2\right)< x^*<1 .Defining G(x) = ax^3+bx^2+cx+d, with a>0, we have{\equation G(-\infty )<0, \quad \quad G\left(-\frac{k^{(U)}-2}{4k^{(I)}+k^{(U)}-2}\right)>0, \quad \quad G(1)<0, \quad \quad G(\infty )>0.\endequation }Therefore, Eq.~(\ref {cubic}) has and has only one solution in the interval (-\left(k^{(U)}-2\right)/\left(4k^{(I)}+k^{(U)}-2\right),1).The other two respectively lie in(-\infty ,-\left(k^{(U)}-2\right)/\left(4k^{(I)}+k^{(U)}-2\right)) and (1,\infty ).\) 
